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MN
5 tháng 3 2020
\(ĐKXĐ:x\ne\pm3\)
\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)
\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)
\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x-3\right)}\)
22 tháng 10 2019
Bài 1:
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+1\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3-x+y\right)\)
\(=2\left(x-y\right)\left(2x+3+y\right)\)
Bài 2:
\(P=\left(3x-1\right)^2+2\left(3x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(3x-1-x-1\right)^2\)
\(=\left(2x-2\right)^2\)(1)
b) Thay \(x=\frac{9}{4}\)vào (1) ta được:
\(\left(2.\frac{9}{4}-2\right)^2\)
\(=\frac{25}{4}\)
Vậy giá trị của P \(=\frac{25}{4}\)khi \(x=\frac{9}{4}\)
Bài 3:
Ta có: \(M=x^2+4x+5\)
\(=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+2\right)^2+1\ge0+1;\forall x\)
Hay \(M\ge1;\forall x\)
Dấu"="xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x=-2\)
Vậy \(M_{min}=1\Leftrightarrow x=-2\)
22 tháng 10 2019
Bài 1 : trên là sai nha mình làm lại
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3y-x+y\right)\)
\(=2\left(x-y\right)\left(2x+4y\right)\)
\(=4\left(x-y\right)\left(x+2y\right)\)
8 tháng 3 2020
ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)
\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)
8 tháng 3 2020
\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)
\(\Leftrightarrow M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)
\(\Leftrightarrow M=\frac{x-9}{2x}\)
Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)
24 tháng 11 2019
Ta có:
a) M = \(\left(\frac{6x}{x^2-9}-\frac{1}{x+3}+\frac{5}{3-x}\right):\frac{4}{x^2-3x}\)
M = \(\left(\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x^2-3x}{4}\)
M = \(\left(\frac{6x-x+3-5x-15}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x\left(x-3\right)}{4}\)
M = \(\frac{-12.x\left(x-3\right)}{\left(x-3\right)\left(x+3\right).4}\)
M = \(-\frac{3x}{x+3}\)
b) Với x = 2 => M = \(-\frac{3.2}{3+2}=-\frac{6}{5}\)
1 tháng 11 2019
a) \(P=\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{2\left(2x+1\right)\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}+\frac{3\left(2x+3\right)\left(2x-3\right)}{\left(2x+1\right)\left(2x+3\right)\left(2x-3\right)}-\frac{\left(6x+5\right)\left(2x+1\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{\left(4x+2\right)\left(2x-3\right)+3\left(4x^2-9\right)-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{8x^2-8x-6+12x^2-27-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{8x^2-24x-38}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
Check hộ mình xem nghi nghi sai sai
1 tháng 11 2019
b) \(Q=\left(\frac{x+1}{2x-1}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2x-1}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{2\left(x+1\right)\left(x-1\right)\left(x+1\right)}{2\left(2x-1\right)\left(x-1\right)\left(x+1\right)}+\frac{2.3\left(2x-1\right)}{2\left(x-1\right)\left(x+1\right)\left(2x-1\right)}-\frac{\left(x+3\right)\left(2x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(2x-1\right)\left(x-1\right)}\right).\frac{4x^2-4}{5}\)
\(=\frac{2\left(x+1\right)\left(x^2-1\right)+12x-6-\left(2x^2+5x-3\right)\left(x-1\right)}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2\left(x^3+x^2-x-1\right)+12x-6-2x^3-5x^2+3x+2x^2+5x-3}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2x^3+2x^2-2x-2+20x-2x^3-3x^2-9}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{-x^2+18x-11}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{-x^2+18x-11}{\left(2x-1\right)}.\frac{2}{5}\)
\(=\frac{-2x^2+36x-22}{5\left(2x-1\right)}\)
15 tháng 12 2018
\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)
15 tháng 12 2018
\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)
\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)
==>Sai đề không mem
25 tháng 8 2020
a) ( x - 5 )( 2x + 3 ) + 2x( 1 - x )
= 2x2 - 7x - 15 + 2x - 2x2
= -5x - 15
= -5( x + 3 )
b) ( 3x - 5 )2 - ( x + 5 )( 5 - x ) - 5/2( -2x )2
= 9x2 - 30x + 25 + ( x + 5 )( x - 5 ) - 5/2.4x2
= 9x2 - 30x + 25 + x2 - 25 - 10x2
= -30x
c) ( 3x + 2 )( 4 - 6x + 9x2 ) - 3x( 3x - 2 )2 + 12( -2/3 - 3x2 )
= ( 3x )3 + 23 - 3x( 9x2 - 12x + 4 ) - 8 - 36x2
= 27x3 + 8 - 27x3 + 36x2 - 12x - 8 - 36x2
= -12x
-=x-5/2x
cái đó bạn rút gọn cho phân thức chung là bao nhiêu zậy